[Math] Why are the trivial zeros of the Riemann zeta function only negative

Question

The functional equation of the Riemann zeta function is
$$\zeta(s)=2^s\pi^{s-1}\sin(s\pi/2)\Gamma(1-s)\zeta(1-s)$$
clearly $2^s$ and $\pi^{1-s}$ are never equal to zero on the complex plane, and neither is Gamma. Some of the zeros can be determined by $\sin(s\pi/2)=0$ but this is the case when $s=2n$ This would imply that there is a zero at every even integer, but it's known that the only non-trivial zeros are at the negative even integers. What am I doing wrong here?

Answer 1

The Riemann zeta function satisfies the functional equation (known as the Riemann functional equation or Riemann's functional equation) $$ \zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s) \!, $$ where Γ(s) is the gamma function, which is an equality of meromorphic functions valid on the whole complex plane. This equation relates values of the Riemann zeta function at the points s and 1 − s. Owing to the zeros of the sine function, the functional equation implies that ζ(s) has a simple zero at each even negative integer $s = −2n$ — these are known as the trivial zeros of ζ(s). When $s$ is an even positive integer, the product $\sin(πs/2)Γ(1−s)$ on the right is regular and non-zero because $Γ(1−s)$ has a simple pole: the functional equation thus relates the values of the Riemann zeta function at odd negative integers and even positive integers.

from Wiki:Riemann functional equation